Problema Solution
Suppose tthat a cyclist began a 261 mi ride across a state at the western edge of the state at the same that a car traveling toward it leaves the eastern end of the state if the bicycle and car met after 4.5 hr and the car traveled 34.2 mph faster than the bicycle find average rate of each
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let
d1 = 2611 mi the distance that the cyclist rode
t = 4.5 hr the time of the travel
v1 = the average rate of the cyclist
v2 = the average rate of the car
the car traveled 34.2 mph faster than the bicycle
v2 = 34.2 + v1
since average rate = distance/time => distance = time*avg.rate
v1*t = d1
v1*4.5 = 2611
by solving we find
v1 = 580.22 mph
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v2 = 34.2 + 580.22
v2 = 614.42 mph
the average rate of the bicycle is 580.22 mph.
the average rate of the car is 614.42 mph.